Can You Fool The Bank With Your Counterfeit Bills?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-size and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint or have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

From Jason Ash, a racket riddle:

You are an expert counterfeiter, and you specialize in forging one of the most ubiquitous notes in global circulation, the U.S. $100 bill. You’ve been able to fool the authorities with your carefully crafted C-notes for some time, but you’ve learned that new security features will make it impossible for you to continue to avoid detection. As a result, you decide to deposit as many fake notes as you dare before the security features are implemented and then retire from your life of crime.

You know from experience that the bank can only spot your fakes 25 percent of the time, and trying to deposit only counterfeit bills would be a ticket to jail. However, if you combine fake and real notes, there’s a chance the bank will accept your money. You have $2,500 in bona fide hundreds, plus a virtually unlimited supply of counterfeits. The bank scrutinizes cash deposits carefully: They randomly select 5 percent of the notes they receive, rounded up to the nearest whole number, for close examination. If they identify any note in a deposit as fake, they will confiscate the entire sum, leaving you only enough time to flee.

How many fake notes should you add to the $2,500 in order to maximize the expected value of your bank account? How much free money are you likely to make from your strategy?

Riddler Classic

From Tyler Barron, a pie puzzle:

You work for Puzzling Pizza, a company committed to making the best and most consistent pizzas in town. As the shop’s resident mathematician, your boss has asked you to design a product similar to one they saw online:

They’ve already purchased the equipment but need to know the exact path and flow rate the sauce-dispensing arm should use to fill a 12-inch circular pizza with sauce as fully and evenly as possible.

Similar to the video, the pizza dough sits on a platform rotating at a constant speed. The sauce-dispensing arm can travel in a straight line across the pizza and distribute sauce in a circular shape with a 0.5-inch diameter. Once the arm starts pouring sauce, it can’t stop until the pizza is done, and the sauce can only be poured onto the dough (i.e., you can’t put sauce on sauce). At the end of your routine, the pizza should have an even layer of sauce on as much of the pizza as possible.

What path and flow rate should the sauce-dispensing arm take to give you the best pizza? (Express the path as the distance from the center of the pizza with respect to angle and the flow rate as the relative flow with respect to the angle.)

Solution to last week’s Riddler Express

Last week brought us to the billiard hall where we were presented with 12 billiard balls. They all looked and felt identical. However, one of the balls was slightly heavier or slightly lighter than the others, but you didn’t know which ball it was or whether it was heavier or lighter. You did, however, have a balance scale which you were allowed to use three times. You could place any equal number of balls on each side of the scale, and the scale would tilt if one side differed in weight. How could you determine which ball was different, and whether it was heavier or lighter?

This puzzle’s submitter, Corey Grodner, described how to accomplish that measuring feat. To begin with — and this is the really important bit — separate the balls into three groups. To keep track of them, label them like so: A1, A2, A3, A4, B1, B2, B3, B4, C1, C2, C3 and C4.

Use your scale a first time to weigh the four balls in the A group against the four balls in the B group. There are three possible scenarios that could result. One, the groups weigh the same. Two, the A balls are heavier. Three, the B balls are heavier. (The third scenario, as far as solving this puzzle goes, is essentially identical to the second scenario, so we’ll just lay out what to do in Scenarios One and Two below.)

Scenario One: Now you know for sure that the abnormal ball is in group C. You also know that all of the A and B balls are normal, and can therefore be used as controls — a handy trick in this billiard hall. Use your scale a second time to weigh C1, C2 and C3 against A1, A2 and A3. If this weighing is equal, you know C4 must be the abnormal ball and you can weigh it against A1 (your third and final use of the scale) to determine if it’s heavier or lighter than normal. If this weighing is unequal, then you know the abnormal ball is C1, C2 or C3. You can weigh C1 against C2 to find out which of the three it is, and you’ll know whether its heavier or lighter based on the results of the weighing that came before.

Scenario Two: Now you know for sure that the abnormal ball is in either group A or B — and moreover that a ball in A is heavier or a ball in B is lighter. You also know that the balls in group C are normal, and can therefore be used as controls. Use your scale a second time to weigh A1, A2, A3 and B1 against A4, C1, C2 and C3. If this weighing is equal, you know that an abnormally light ball is one of B2, B3 and B4, and you can weigh B2 against B3 (your final use) to figure out which. If that second weighing shows that the left side is heavier, you know one of A1, A2 and A3 is the culprit and you can weigh A1 against A2 to suss it out. If that second weighing shows that the right side is heavier, you know that either B1 is abnormally light or that A4 is abnormally heavy. Weigh B1 against C1 to figure it out.

Phew! Congratulations on ferreting out that pesky abnormal ball. And now that you’ve performed your three measurements, can I have my scale back, please?

Solution to last week’s Riddler Classic

Last week, we reintroduced a two-player map-coloring game. It worked like this: Call the two players Allison and Bob. On each turn, Allison draws a simple closed curve on a piece of paper. Bob then colors the interior of the “country” that curve creates with one of his many crayons. If the new country borders any existing countries, Bob must color the new country with a color different from the ones he used for the bordering countries. In last week’s edition of the game, Allison won when she forced Bob to use a seventh color. How many countries did Allison have to draw to win? (In an earlier version, only six colors were necessary for victory, and Allison had to draw eight countries to achieve that.)

If Bob is playing optimally, only using a new color if he must, Allison had to draw 12 countries to win.

Allison begins by creating a chain of countries, forcing Bob to use two colors, which alternate. After that chain is five countries long, Allison draws a country which borders two countries in the chain, forcing Bob to use a third color. She then draws a country which borders that country, as well as two countries in the chain, forcing Bob to use a fourth color. And so on. As the countries continue to overlap, of course, Bob is further forced to pull new crayons out of his box. Our winner this week, Zach, was kind enough to share an image of his work, showing that winning arrangement.

There does not appear to be a limit to how many colors Allison can eventually force Bob to use, and a pattern like the one above can grow indefinitely. Sadly, the largest box of Crayolas appears to contain only 120 different colors.

Want more riddles?

Well, aren’t you lucky? There’s a whole book full of the best puzzles from this column and some never-before-seen head-scratchers. It’s called “The Riddler,” and it’s in stores now!

Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.

Footnotes

Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. Eastern time on Sunday. Have a great weekend!